Originally Posted by

**ThePerfectHacker** No, I am not familar with this theorem. I do not even understand the notation you are using.

However, I did think of an elementary way of proving $\displaystyle \text{kum}(K/F) = \left< \sqrt[n]{a_1}F^{\times}, ... , \sqrt[n]{a_r}F^{\times}\right>$. We note that every element in $\displaystyle K = F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ has to be expressed in terms of powers and linear combinations of $\displaystyle \sqrt[n]{a_i}$'s. For example, consider $\displaystyle K=F(\sqrt{2},\sqrt{3})$ (here $\displaystyle F=\mathbb{Q}$) then $\displaystyle \{1,\sqrt{2}\}$ is basis for $\displaystyle F(\sqrt{2})/F$ and $\displaystyle \{1,\sqrt{3}\}$ is basis for $\displaystyle F(\sqrt{2},\sqrt{3})/F(\sqrt{2})$. Thus, basis for $\displaystyle K/F$ is $\displaystyle \{1,\sqrt{2},\sqrt{3},\sqrt{2} \cdot \sqrt{3}\}$. Therefore, any element can be written as $\displaystyle a+b\sqrt{2}+c\sqrt{3}+d\sqrt{2}\cdot \sqrt{3}$. And this is true in general, any element in a Kummer extension can be written out in terms of linear combinations of the products of the $\displaystyle n$-th roots. Let $\displaystyle \{x_1,...,x_m\}$ be a basis for $\displaystyle K= F(\sqrt[n]{a_1},...,\sqrt[n]{a_r})$ where $\displaystyle x_i$'s are products of the $\displaystyle \sqrt[n]{a_j}$'s by above. If $\displaystyle \alpha \in K$ then we can write $\displaystyle \alpha = f(x_1,...,x_m)$ where $\displaystyle f$ is a polynomial over $\displaystyle F$ with $\displaystyle m$ variables. Assume that $\displaystyle \alpha$ has the property that $\displaystyle \alpha^n \in F$. We wish to argue that __all__ terms of $\displaystyle f(x_1,...,x_m)$ must be zero except for possibly one of them, and this will show that the only elements in the Kummer group are the products of the $\displaystyle n$-th roots. Suppose this was not the case and there were two coefficients for $\displaystyle x_i,x_j$ which were non-zero. Then in the expansion of $\displaystyle \alpha^n = f^n (x_1,...,x_m)$ we would have a non-zero "cross term" $\displaystyle x_ix_j$. But then this means $\displaystyle \alpha^n \not \in F$ by the uniqueness of the representation of an element in terms of a basis.